mateforum.ro

Posted: **Wed Nov 19, 2008 8:49 pm**

Fie \( a , b , c \in \mathb{R}_+^* \) . Aratati ca :
\( a\sqrt{{b+c}}+b\sqrt{c+a}+c\sqrt{a+b}\le\sqrt{2(a+b+c)(ab+bc+ca)}; \)

\( \frac{a}{sqrt{b+c}}+\frac{b}{sqrt{c+a}}+\frac{c}{sqrt{a+b}}\ge(a+b+c)sqrt{\frac{a+b+c}{2(ab+bc+ca}} \)

Cezar Lupu

Posted: **Wed Nov 19, 2008 11:17 pm**

\( LHS=\sum{\sqrt{a}\sqrt{a(b+c)}}\le\sqrt{(a+b+c)[a(b+c)+b(c+a)+c(a+b)]}=RHS \)

Posted: **Wed Nov 19, 2008 11:30 pm**

\( LHS=\sum{\frac{a^2}{a{\sqrt{b+c}}}\ge\frac{(a+b+c)^2}{\sum{a\sqrt{b+c}}}\ge\frac{(a+b+c)^2}{\sqrt{2(a+b+c)(ab+bc+ca)}}=RHS \)

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Constanta 2006

Constanta 2006